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I‘m taking a course in functional analysis and have to show that $ L^1(\mathbb R) \subsetneq (L^\infty(\mathbb R))^* $ "in the sense of canonical embedding". What does this mean exactly? Unfortunately, "canonical embedding" is no term we defined in the lecture so I assume it is some "common term".

I think I have to show that there is a function

$$ \iota: L^1(\mathbb R) \to (L^\infty(\mathbb R))^* $$

such that $ \iota(L^1(\mathbb R)) \subsetneq (L^\infty(\mathbb R))^* $. But just any function is probably not enough and $\iota$ has to satisfy some more properties. For example, i think injectivity would make sense.

Can somebody tell me what one means with a "canonical embedding"? Is it just a topological embedding, i.e. a homeomorphism onto its image as it is mentioned at Wikipedia?

ATW
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  • A canonical embedding is an embedding that is understood from the context (that has been previously agreed upon to be canon). So here i would understand this to be the map that associates to an $L^1$ function an element in $(L^\infty)^*$ by integration against that function. So you need to show that this specific map is not surjective. – jd27 Jun 21 '23 at 17:26
  • @jd27 But is just a map enough? Then choose $\iota$ as the constant null function and we’re done. This can’t be the intended solution. – ATW Jun 21 '23 at 17:38
  • i wrote "specific" map, how do you get the idea that this is about any map? See the answer from Trevor Gunn for the specific map i mean. – jd27 Jun 21 '23 at 17:40
  • Due to Hölder's inequality, there's a natural (i.e. canonical) way to begin with $g \in L^q(\mathbb{R})$ and produce a functional $\kappa_p(g) \in (L^p(\mathbb{R}))^*$ via integration: $$ \kappa_p(g)(f) = \int_{\mathbb{R}} f(x)g(x) , dx $$ for any $f \in L^p(\mathbb{R})$. This holds whenever $\frac1p + \frac1q = 1$, including the case $(p, q) = (\infty, 1)$. – Sammy Black Jun 21 '23 at 17:40
  • Concerning the word "canonical" see https://mathoverflow.net/q/19644 and https://mathoverflow.net/q/353061. – Paul Frost Jul 26 '23 at 07:58

1 Answers1

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"Canonical" means the map defined most directly by the definitions/properties. Meaning there's no choice of basis or any elements being permuted. Here, this map is defined by the pairing

\begin{align} L^1(\mathbb{R}) \times L^\infty(\mathbb{R}) &\to \mathbb{R} \\ (f, g) &\mapsto \int_{\mathbb{R}} fg \end{align}

Every element of $L^1$ gives a functional $f^* = (f, -)$ on $L^\infty$ and that's the map: $f \mapsto f^*$.

Trevor Gunn
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  • And embedding means what? Just inclusion? – ATW Jun 21 '23 at 17:34
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    @ATW Yes, an injective map – Trevor Gunn Jun 21 '23 at 17:43
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    With sets, we speak of an injective map, but when the sets carry extra structure (hence are called spaces) then it's more common to speak of embedding as long as the injection respects this extra structure. In this case, $L^p(\mathbb{R})$ is a vector space, and the canonical dual map is linear, i.e. respects the vector space structure. – Sammy Black Jun 21 '23 at 17:48